I tried to solve a question which led to
$$4 \sin\frac{\pi}{5} \sin\frac{3\pi}{5} = \sqrt{5}$$
But, I got stuck to prove that.
Is there any easy solution?
$\endgroup$ 14 Answers
$\begingroup$Note that
$$\sin{\frac{3 \pi}{5}} = \sin{\frac{2 \pi}{5}}$$
then use a double-angle and triple-angle forumla:
$$\sin{2 x} = 2 \sin{x} \cos{x}$$ $$\sin{3 x} = 3 \sin{x} - 4 \sin^3{x}$$
In this case, $x=\pi/5$. Setting the above two equations equal to each other results in the quadratic equation in question:
$$2 \cos{x} = 3 - 4 (1-\cos^2{x}) = 4 \cos^2{x}-1$$
which means that
$$\cos{x} = \frac{1 + \sqrt{5}}{4}$$ $$\sin{x} = \frac{\sqrt{10-2\sqrt{5}}}{4}$$
Then, using $\sin{2 x} = \sin{3 x}$ as noted above, we have
$$4 \sin{x} \sin{2 x} = 8 \sin^2{x} \cos{x} = 8 \frac{10-2 \sqrt{5}}{16} \frac{1+\sqrt{5}}{4} = \frac{10-10-2 \sqrt{5}+10\sqrt{5}}{8} = \sqrt{5}$$
$\endgroup$ 1 $\begingroup$Using this,
$\sin5x=16\sin^5x-20\sin^3x+5\sin x$
Let $\sin5x=0\implies 5x=n\pi$ where $n$ is any integer $\displaystyle\implies x=\frac{n\pi}5$
Clearly, $\displaystyle\sin\frac{\pi}5, \sin\frac{3\pi}5$ satisfy $16\sin^5x-20\sin^3x+5\sin x=0$
Again, if $\sin x=0,x=m\pi$ where $m$ is any integer
$\implies \displaystyle\sin\frac{\pi}5, \sin\frac{3\pi}5$ satisfy $16\sin^4x-20\sin^2x+5=0$
$\displaystyle\implies \sin^2x=\frac{20\pm\sqrt{20^2-4\cdot16\cdot5}}{2\cdot16}=\frac{5\pm\sqrt5}8$
Now, as $\displaystyle0<\frac\pi5,\frac{3\pi}5<\pi;\sin\frac{\pi}5, \sin\frac{3\pi}5>0,$ the values of $\displaystyle \sin\frac{\pi}5, \sin\frac{3\pi}5 $ will be $\displaystyle\sqrt{\frac{5\pm\sqrt5}8}$
$\displaystyle\implies \sin\frac{\pi}5\cdot\sin\frac{3\pi}5=\sqrt{\frac{5+\sqrt5}8}\cdot\sqrt{\frac{5-\sqrt5}8}=\frac{\sqrt{(5+\sqrt5)(5-\sqrt5)}}8=\frac{\sqrt5}4$
$\endgroup$ 3 $\begingroup$I will find the $\cos\frac{2\pi}{5}$ with basic complex number theory.
Let $z = \cos(\frac{2\pi}{5}) + i \sin(\frac{2\pi}{5})$, then $z^5=1$. (de Moivre's Theorem) Let's find the solution of this quintic equation.
This equation is factored into $(z-1)(z^4 + z^3 + z^2 + z +1 ) = 0$. $z=1$ is trivial solution. Now let $u = z + z^4$, and $v= z^2 + z^3$. then $u + v = -1$, and $uv = z^3 + z^4 + z^6 + z^7 = z + z^2 + z^3 + z^4 = -1$. So $t^2 + t - 1 = 0$ has two solutions $u$ and $v$. Solving the quadratic equation, we get $t = \frac{-1 \pm \sqrt{5}}{2}$.
Geometrically $u$ has positive real part, and $v$ has negative real part. so $z + z^4 = \frac{-1 + \sqrt{5}}{2}$. Since $z$ and $z^4$ are conjugate, the real part of $z$ is the half of it, i.e. $\cos\frac{2\pi}{5}=\frac{-1 + \sqrt{5}}{4}$.
Now, $\cos(\frac{2\pi}{5}) = 1-2\sin^2(\frac{\pi}{5})$, so you can find $\sin\frac{\pi}{5}$.
In this way you can find $\cos\frac{2\pi}{17}$, by decreasing the order of equation 16th to 8th, 4th, .... You need to take $u$ and $v$ in special way...
$\endgroup$ 3 $\begingroup$Using $2\sin A\sin B=\cos(A-B)-\cos(A+B),$
$2\sin36^\circ\sin108^\circ=\cos72^\circ-\cos144^\circ>0$
As $\frac{144-72}2=36,$
$\displaystyle\cos72^\circ+\cos144^\circ=\frac{2\sin36^\circ\cos72^\circ+2\sin36^\circ\cos144^\circ}{2\sin36^\circ}$
Using $2\sin B\cos A=\sin(A+B)-\sin(A-B),$
$\displaystyle \frac{2\sin36^\circ\cos72^\circ+2\sin36^\circ\cos144^\circ}{2\sin36^\circ}$ $\displaystyle=\frac{\sin108^\circ-\sin36^\circ+2\sin180^\circ-\sin72^\circ}{2\sin36^\circ}=-\frac12$ as $\sin108^\circ=\sin(180^\circ-72^\circ)=\sin72^\circ$
$\displaystyle\cos72^\circ+\cos144^\circ=-\frac12\ \ \ \ (1)$
Again using $\sin2A=2\sin A\cos A$, $\displaystyle\cos72^\circ\cos144^\circ=\frac{(2\sin72^\circ\cos72^\circ)\cos144^\circ}{2\sin72^\circ} =\frac{\sin144^\circ\cos144^\circ}{2\sin72^\circ}$ $\displaystyle=\frac{2\sin144^\circ\cos144^\circ}{2\cdot2\sin72^\circ}=\frac{\sin288^\circ}{4\sin72^\circ}=\frac{\sin(360^\circ-72^\circ)}{4\sin72^\circ}=\frac{-\sin72^\circ}{4\sin72^\circ}=-\frac14\ \ \ \ (2)$
$\displaystyle\implies \cos72^\circ-\cos144^\circ=+\sqrt{(\cos72^\circ+\cos144^\circ)^2-4\cos72^\circ\cos144^\circ}$
$\displaystyle =+\sqrt{\left(-\frac12\right)^2-4\left(-\frac14\right)}$ using $(1),(2)$
$\displaystyle =\frac{\sqrt5}2$
$\displaystyle\implies 2\sin36^\circ\sin108^\circ=\frac{\sqrt5}2$
Alternatively, using this or this, $\displaystyle\cos36^\circ=\frac{\sqrt5+1}4$
Use $\cos2A=2\cos^2A-1$ to find $\cos72^\circ=\frac{\sqrt5-1}4$
and $\cos144^\circ=\cos(180^\circ-36^\circ)=-\cos36^\circ$
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